2020 AMC 8 Problems/Problem 17

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Video Solution by NiuniuMaths (Easy to understand!)
6 Video Solution by Math-X (First understand the problem!!!)
7 Video Solution (🚀Just 3 min🚀)
8 Video Solution by OmegaLearn
9 Video Solution by North America Math Contest Go Go
10 Video Solution by WhyMath
11 Video Solution
12 Video Solution by Interstigation
13 See also

Problem

How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$ )

$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution 1

Since $2020 = 2^2 \cdot 5 \cdot 101$ , we can simply list its factors: $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.$ There are $12$ factors; only $1, 2, 4, 5, 101$ don't have over $3$ factors, so the remaining $12-5 = \boxed{\textbf{(B) }7}$ factors have more than $3$ factors.

Solution 2

As in Solution 1, we prime factorize $2020$ as $2^2\cdot 5\cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number which has one factor is $1$ . For a number to have exactly two factors, it must be prime, and the only prime factors of $2020$ are $2$ , $5$ , and $101$ . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of $2020$ is $4$ . Thus, there are $5$ factors of $2020$ which themselves have $1$ , $2$ , or $3$ factors (namely $1$ , $2$ , $4$ , $5$ , and $101$ ), so the number of factors of $2020$ that have more than $3$ factors is $12-5=\boxed{\textbf{(B) }7}$ .

Solution 3

Let $d(n)$ be the number of factors of n. We know by prime factorization that $d(2020) = 12$ . These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2020$ . Since $d(2020) = d(a)d(b)$ , one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2020$ with more than $3$ factors. However, this argument has exceptions where $a$ and $b$ share a nontrivial common factor, which in this case can only be two. There are two cases - One in which $5$ and $101$ divide the same factor, WLOG assumed to be $a$ , so that $d(a) = 2^3 > 3$ and $d(b) = 2$ , as otherwise. In the other case, $a = 5\cdot2$ and $b = 101\cdot2$ , so that $d(a) = d(b) = 4$ . This adds one factor with more than $3$ factors, so the answer is $\boxed{\textbf{(B) }7}$ .

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=YsWfaht72aFTG3eZ&t=3020

~Math-X

Video Solution (🚀Just 3 min🚀)

https://youtu.be/jJnxvFJuhQw

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/Of-ZGiWgXyY?t=56

~ pi_is_3.14

Video Solution by North America Math Contest Go Go

https://www.youtube.com/watch?v=tUTFLUJ6a-4

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/2dazhQ31I14

~savannahsolver

Video Solution

https://youtu.be/VnOecUiP-SA

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=782

~Interstigation

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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